Operator norm on a Hilbert module

Question

Let $$ be a (right) Hilbert module over a $C^*$-algebra $$. Does it always hold that the $C^*$-norm of $a \in $ is equal to its operator norm, when we consider $a$ as a bounded operator on $X$, or can the norm be strictly less? If yes, do such Hilbert modules have a name?

Answer 1

What you're essentially asking is whether or not the $*$-homomorphism $A\to B_A(X)$ is always isometric, or equivalently, injective. As stated in the comments, this is not true for every Hilbert $C^*$-module. I don't know if this property has a name, and it isn't given a name in any of the references on Hilbert $C^*$-modules I have found.

Note that if $X$ is a full Hilbert $A$-module, i.e., the ideal $$\langle X,X\rangle=\operatorname{span}\{\langle x,y\rangle\in A\mid x,y\in X\}$$ is dense in $A$, then $X$ has this property. Indeed, if we let $$N(X)=\{a\in A\mid xa=0\text{ for all }x\in X\},$$ then we have $N(X)\overline{\langle X,X\rangle}=N(X)\cap\overline{\langle X,X\rangle}=0$, and the map $A\to B_A(X)$ is injective if and only if $N(X)=0$.

Unforunately this fails to be an equivalence, for example $A=B(H)$ where $H$ is a separable infinite-dimensional Hilbert space, and $X=H$. In this case clearly the map $B(H)\to B(H)$ is the identity (hence injective), but $\langle X,X\rangle=K(H)$ is the compact operators.